Suppose that $C^*_r(G)\cong C^*_r(H)$, can we conclude that $G\cong H$?
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4$\begingroup$ Take two non-isomorphic finite groups $G, H$ with isomorphic group algebras. $\endgroup$– JCACommented May 29, 2020 at 2:43
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$\begingroup$ In general the conclusion does not hold, as pointed out by JCA. But some classes of groups have the property you are asking for. Such groups are usually called C*-superrigid. $\endgroup$– worldreporterCommented May 29, 2020 at 17:58
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The simplest counterexample is $G = \mathbb{Z}_2\times \mathbb{Z}_2$ and $H = \mathbb{Z}_4$. These groups are not isomorphic but $C^*(G) \cong \mathbb{C}^4 \cong C^*(H)$.
answered May 29, 2020 at 2:56
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$\begingroup$ Pro Nik weaver, May I ask you a question: You proved that calkin algebra has outer automorphisms. Does there exist other open questions concerning $C^*$-algebras which have automorphisms? $\endgroup$ Commented May 30, 2020 at 14:46
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$\begingroup$ @mathbeginner: nothing comes to mind. $\endgroup$ Commented May 30, 2020 at 17:21